document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=11826>Question 11826</A>: <I><SPAN STYLE=\"word-break: break-all;\"> hi im having trouble on this problem. \"graph the line perpendicular to the graph of 3x-2y=24 that intersects it at its x-intercepts.\" what i did was get the equation into the yintercept form i think its called like that. y=-12+3/2x. the x-intercept is 3/2 right? and i plot the -12 on the graph right? now i dont know what to do. thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #6056 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=6056\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> no, the straight line in \"y-intercept\" form or whatever you want to call it is:\r<BR>\n" );
document.write( "\n" );
document.write( "y = (3/2)x - 12... always think of it as y = mx + c ie x term first.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so, the gradient is 3/2 and the y-intercept is -12.<BR>\n" );
document.write( "If you are going to plot this properly, then we need 2 points to join. We now have (0, -12).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The second point we can find easily is the x-intercept, when y=0...<BR>\n" );
document.write( "3x-2y=24<BR>\n" );
document.write( "3x = 24<BR>\n" );
document.write( "x = 24/3<BR>\n" );
document.write( "--> x=8<BR>\n" );
document.write( "--> so the point is (8,0)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "This allows you to plot the first equation. How about plotting the second, the equation that is perpendicular?\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "well, it will fit the line y=mx+c again. We need its gradient and y-intercept.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Its gradient is -2/3... 2 lines perpendicular to each other will have their gradients m and n where mn=-1. This is ALWAYS true\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "(eg if one gradient is 3, the other will be -1/3 etc)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so, we have y = -(2/3)x + c\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "To find c, we need to know a (x,y) pair, which we do, because we know the line goes through the point (8,0), so 0 = -(2/3)(8) + c<BR>\n" );
document.write( "-16/3 + c = 0<BR>\n" );
document.write( "c = 16/3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "--> equation is y = -(2/3)x + 16/3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "now to find another point on this line, to draw the straight line... pick the easiest one you can, as ever... so when x=0, y = 16/3, so now you can plot the line.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "jon.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );